var sidebar_align = 'right';
var content_container_margin = parseInt('290px');
var sidebar_width = parseInt('270px');
Unanswered: "char" as a table name causing problems in PHP code...
Every time I execute my code, it returns the error
$result = mysql_query("SELECT * FROM char",$db);
However when I change "char" to "login", it works fine...
Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in C:\htdocs\index.php on line 37
All I can figure is causing it is that the table is named "char" which, I'm assuming, is a restricted word...
I searched all over Google and couldn't figure out what to look for, then tried replacing char with a variable ".$chardb." which is defined as ... Which gave the same compilation error.
So, my question is, besides changing the name of the table, what can I do to make this work?
Thanks in advance
i do that all the time too
Originally Posted by L33tN3k0
try putting the table name in backticks --
select foo from `char`